Binomial Theorem Practical Problem

Question

I have been studying the 'Binomial Series', Chapter 16, Pg.125 within the Engineering Mathematics Book by John Bird. After completing this section I have attempted to complete the exercises for practical problems (10 & 12) involving the binomial theorem (pg.133).

I have struggled to complete the following question. I have been suggested to use logarithms but can one not substitute values in for the percentages where it can then be rearranged?

A first step, or point of direction/suggestion would be brilliant. Thank You.

The viscosity $\eta $ of a liquid is given by; $$ \eta = \frac{(kr)^4}{vl} $$

Where $k$ is a constant.

If there is an error in $r$ of $+2\%$, in $v$ of $+4\%$ and $l$ of $-3\%$, what is the resultant error in $\eta$?

Answer 1

So the true quantity is $r$, but the measured quantity is $r'=(1+2\%)r$ and similarly for the other quantities?

Yes, you can insert the exact formulas to get

$$η'=η\cdot\frac{(1+2\%)^4}{(1+4\%)(1-3\%)}$$

but quick'n'dirty you get in first order a cumulative error of

$$4\cdot 2\%-1\cdot 4\%-1\cdot(-3\%)=7\%$$

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The expression is of the form

\begin{align} & (1+\alpha)^4(1+\beta)^{-1}(1-\gamma)^{-1} \\[10pt] = {} & (1+4α+6α^2+4α^3+α^4)\,(1-β+β^2-β^3+β^4+\cdots)\,(1+γ+γ^2+γ^3+γ^4+\cdots) \end{align}

and by multiplying out starting from the lowest degree terms one gets

$$1+(4α-β+γ)+(6α^2-4αβ+β^2+4αγ-βγ+γ^2)+\cdots $$

With that one now has to decide if there is still an essential contribution from the quadratic or higher order terms, in this case there is not.

Answer 2

$$\eta = \frac{(kr)^4}{vl}=\frac{k^4r^4}{vl}$$

$$\text{taking $\ln$ on both sides} \implies \begin{align}\ln \eta&=\ln(k^4r^4)-\ln(vl)\\ &=4\ln k+4\ln r-\ln v-\ln l\\ \end{align}$$

$$\text{differentiate w.r.t $\eta$}\implies{1\over \eta}={4\over k}\require{cancel}\cancelto{0\; \because\ k\; constant}{{{\partial k}\over{\partial \eta}}}+{4\over r}{{\partial r}\over{\partial \eta}}-{1\over v}{{\partial v}\over{\partial \eta}}-{1\over l}{{\partial l}\over{\partial \eta}}$$

$$\text{For sufficiently small ${\partial x}\over \partial y$} \text{,$\quad {{\partial x}\over \partial y}\approx\frac{\Delta x}{\Delta y}$}$$

$${\rm Hence} {1\over \eta}={4\over r}{{\Delta r}\over{\Delta \eta}}-{1\over v}{{\Delta v}\over{\Delta \eta}}-{1\over l}{{\Delta l}\over{\Delta \eta}}$$

$${\rm {or\ equivalently;\quad}}{{\Delta \eta}\over \eta}={{4\Delta r}\over{r}}-{{\Delta v}\over{ v}}-{{\Delta l}\over{l}}\\$$ Hence (since fractional error in $x={{\Delta x}\over x}$), (fractional/percentage) error in $\eta$=

$${{\Delta \eta}\over \eta}=4(0.02)-0.04-(-0.03)=0.07=7\%$$